1. Stating the problem: We want to find how many kilograms of 6% solution must be mixed with 3 kg of 9% solution to get a 7% solution. 2. Formula: Let $x$ be the kg of 6% solution. The total amount of solute before mixing equals the total amount of solute after mixing. $$0.06x + 0.09 \times 3 = 0.07(x + 3)$$ 3. Solve the equation: $$0.06x + 0.27 = 0.07x + 0.21$$ 4. Rearrange terms: $$0.27 - 0.21 = 0.07x - 0.06x$$ $$0.06 = 0.01x$$ 5. Divide both sides by 0.01: $$\cancel{0.06} \div \cancel{0.01} = \cancel{0.01x} \div \cancel{0.01}$$ $$6 = x$$ 6. Answer: $6$ kg of 6% solution is needed. --- 1. Stating the problem: We want to find how many liters of 5% milk must be mixed with 80 liters of 2% milk to get milk with 2.6% fat. 2. Formula: Let $y$ be the liters of 5% milk. Total fat before mixing equals total fat after mixing. $$0.02 \times 80 + 0.05y = 0.026(80 + y)$$ 3. Calculate: $$1.6 + 0.05y = 2.08 + 0.026y$$ 4. Rearrange terms: $$0.05y - 0.026y = 2.08 - 1.6$$ $$0.024y = 0.48$$ 5. Divide both sides by 0.024: $$\cancel{0.024y} \div \cancel{0.024} = 0.48 \div 0.024$$ $$y = 20$$ 6. Answer: $20$ liters of 5% milk is needed.